The ministers of the cabinet were quite upset by the message from the Chief of Security stating that they would all have to change the four-digit room numbers on their offices.
— It is a matter of security to change such things every now and then, to keep the enemy in the dark.
— But look, I have chosen my number 1033 for good reasons. I am the Prime minister, you know!
— I know, so therefore your new number 8179 is also a prime. You will just have to paste four new digits over the four old ones on your office door.
— No, it’s not that simple. Suppose that I change the first digit to an 8, then the number will read 8033 which is not a prime!
— I see, being the prime minister you cannot stand having a non-prime number on your door even for a few seconds.
— Correct! So I must invent a scheme for going from 1033 to 8179 by a path of prime numbers where only one digit is changed from one prime to the next prime.
Now, the minister of finance, who had been eavesdropping, intervened.
— No unnecessary expenditure, please! I happen to know that the price of a digit is one pound.
— Hmm, in that case I need a computer program to minimize the cost. You don't know some very cheap software gurus, do you?
— In fact, I do. You see, there is this programming contest going on... Help the prime minister to find the cheapest prime path between any two given four-digit primes! The first digit must be nonzero, of course. Here is a solution in the case above.
1033
1733
3733
3739
3779
8779
8179
The cost of this solution is 6 pounds. Note that the digit 1 which got pasted over in step 2 can not be reused in the last step – a new 1 must be purchased.
Input
One line with a positive number: the number of test cases (at most 100). Then for each test case, one line with two numbers separated by a blank. Both numbers are four-digit primes (without leading zeros).
Output
One line for each case, either with a number stating the minimal cost or containing the word Impossible.
Sample Input
3
1033 8179
1373 8017
1033 1033
Sample Output
6
7
0
沒有impossible也過了，。。。。emmmmm。
我的方法很蠢。
先打表，然後一個個搜。
#include <iostream>
#include <stdio.h>
#include<queue>
#include<string.h>
using namespace std;
struct e
{
int data;
int ans;
};
int a[10000],flag;
bool vis[10000];
void da()
{
int i;
for(i=2;i<10000;i++)
{
if(a[i]==0)
{
for(int j=i*2;j<10000;j+=i)
a[j]=1;
}
}
}
void bfs()
{
flag=0;
e e1,e2;
int x,y,ans=0;
scanf("%d%d",&x,&y);
e1.data=x,e1.ans=0;
vis[x]=true;
queue<e>q;
q.push(e1);
while(q.size())
{
e1=q.front();
q.pop();
if(e1.data==y)
{
flag=1;
printf("%d\n",e1.ans);
break;
}
int k=e1.data/10*10;
for(int i=k;i<k+10;i++)
if(!a[i]&&vis[i]==false)
{
e2.data=i;
e2.ans=e1.ans+1;
vis[i]=true;
q.push(e2);
}
k=(e1.data/10)%10;
k=e1.data-k*10;
for(int i=k;i<k+100;i+=10)
if(!a[i]&&vis[i]==false)
{
e2.data=i;
e2.ans=e1.ans+1;
vis[i]=true;
q.push(e2);
}
k=(e1.data/100)%10;
k=e1.data-k*100;
for(int i=k;i<k+1000;i+=100)
if(!a[i]&&vis[i]==false)
{
e2.data=i;
e2.ans=e1.ans+1;
vis[i]=true;
q.push(e2);
}
k=(e1.data/1000)%10;
k=e1.data-(k-1)*1000;
for(int i=k;i<10000;i+=1000)
if(!a[i]&&vis[i]==false)
{
e2.data=i;
e2.ans=e1.ans+1;
vis[i]=true;
q.push(e2);
}
}
if(!flag)printf("Impossible\n");
}
int main()
{
int n,i ;
scanf("%d",&n);
da();
while(n--)
{
memset(vis,false,sizeof(vis));
bfs();
}
// for(i=2; i<10000; i++)
// if(!a[i])
// cout<<i<<endl;
//cout << "Hello world!" << endl;
return 0;
}